The interionic dist in FCC=edge/square root of 2.But in NaCl (FCC) dist between Na+ & Cl- is a/2

Asked by  | 9th Apr, 2008, 10:58: AM

Expert Answer:

Anions ccp (fcc). Radius Na+ = 1.02Å, radius Cl- = 1.81Å; radius ratio = 0.563.

Therefore Na octahedral.

1 octahedral / anion therefore 100% octahedral sites are filled.

Coordination # Na = 6; coordination # Cl = 6. 
For an fcc lattice there are 4 lattice points per cell, the motif in this case is a Na cation and a Cl anion.  Therefore the cell contents are 4 Na cations + 4 Cl anions.

Lattice parameter and Density calculations.

The cation and its nearest neighbor anions are, by definition in contact.
Therefore the cell edge  = 2r(Na+) + 2r(Cl-) = 2(1.02) + 2(1.81) = 5.66Å

The Coulombic interactions involve all the surrounding ions:

  6 Cl-    at a distance    d

12 Na+    at a distance   d√2

  8 Cl-   at a distance   d√3

  6 Na+    at a distance   d√4 = 2d

24 Cl-    at a distance   d√5

                                                      and so on…

Thus we have

     UC  =  Z+Z-e2 ( 6 – 12/√2 + 8/√3 – 6/√4 + 24/√5 - ...)

                                                               4peod

The infinite series can be summed, and Madelung’s constant for the NaCl structure type  MNaCl = 1.748.

 

Answered by  | 20th May, 2008, 09:09: AM

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